Difference between revisions of "1993 AJHSME Problems/Problem 1"

(Solution 2)
 
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==Solution 2==
 
==Solution 2==
We know that if we want a product of 36, both numbers have to be positive or negative. Scanning the number pairs, the only choice with one negative number and one positive number is <math>\boxed{\text{(C)}}</math>  
+
We know that if we want a product of 36, both numbers have to be positive or negative. Scanning the number pairs, the only choice with one negative number and one positive number is <math>\boxed{\text{(C)}}</math>
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~fn106068
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==See Also==
 
==See Also==
 
{{AJHSME box|year=1993|before=First<br />Question|num-a=2}}
 
{{AJHSME box|year=1993|before=First<br />Question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:55, 23 July 2021

Problem

Which pair of numbers does NOT have a product equal to $36$? $\text{(A)}\ \{-4,-9\}\qquad\text{(B)}\ \{-3,-12\}\qquad\text{(C)}\ \left\{\frac{1}{2},-72\right\}\qquad\text{(D)}\ \{ 1,36\}\qquad\text{(E)}\ \left\{\frac{3}{2},24\right\}$

Solution 1

A. The ordered pair ${-4,-9}$ has a product of $-4\cdot-9=36$

B. The ordered pair ${-3,-12}$ has a product of $-3\cdot-12=36$

C. The ordered pair ${1/12, -72}$ has a product of $-36$

D. The ordered pair ${1, 36}$ has a product of $1\cdot36=36$

E. The ordered pair ${3/2, 24}$ has a product of $3/2\cdot24=36$

Since C is the only ordered pair which doesn't equal 36, $\boxed{\text{(C)}}$ is the answer.

Solution 2

We know that if we want a product of 36, both numbers have to be positive or negative. Scanning the number pairs, the only choice with one negative number and one positive number is $\boxed{\text{(C)}}$ ~fn106068

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
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